Ytukyg

Show that function $J$ is semi-continuous at point $vin U$ iff $forall epsilon >0$ there exists $delta >0$ such that $forall uin Ucap L_delta (v)$, $J(u)>J(v)-epsilon $.

$L_delta (v)$ is open ball with center in $v$ and radius $delta$.

Definition of semi-continuity is "Function $J: Urightarrow R, Usubset R^n$ is semi-continuous from under at point $vin U$ if $limlimits_{k to infty} inf J(u_k)geq J(v)$ for every sequence ${u_k}_{kin N}$ that converges to $v$".

Please, can anyone help me to show this.

"If" part: Suppose that the "$varepsilon/delta$" condition holds, and let $u_k$ be a sequence as in definition of semicontinuity. Fix $varepsilon$, and take $delta$ as in the condition. Then for some $n_0$, for $k > n_0$ we have $u_k in L_delta(u)$. Hence $J(u_k) > J(v) - varepsilon$. Passing to $liminf_k J(u_k) geq J(v) - varepsilon$. Since $varepsilon > 0$ was arbitrary, you get $liminf_k J(u_k) geq J(v)$

"Only if" part:
Suppose that $J$ is semicontinuous from below. If the "$varepsilon/delta$" condition was false, then for some $varepsilon > 0$ we would be able to find $u_k$ with $|u_k - v| < frac{1}{k}$ but $J(u_k) leq J(v) - varepsilon$ (else, $delta = 1/k$ would work). Passing to the limit we find $liminf_k J(u_k) leq J(v) - varepsilon$, contradicting semicontinuity.